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Maru [420]
4 years ago
12

A grinding wheel is spinning with an initial angular velocity +ω0. When its motor is turned off at t=0, it begins to slow down w

ith an angular acceleration −α. Derive an expression for the number of revolutions that the wheel will rotate through before coming to rest. Express your answer in terms of the variables ω0 and α, and the constant π.
Physics
1 answer:
Olegator [25]4 years ago
3 0

Answer:

Explanation:

Given

initial angular velocity is \omega _0

when motor is turned of it started decelerating with -\alpha  angular acceleration

angle turned before coming to halt

\omega ^2-\omega _0^2=2(\alpha )\theta

here final angular velocity is zero

-\omega _0^2=-2\alpha \theta

\theta =\frac{\omega _0^2}{2\alpha }

No of revolutions will be N=\frac{\theta }{2\pi }

N=\frac{\omega _0^2}{4\pi \alpha }

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48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers ha
Sedbober [7]

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

Fp=\frac{R_{in}}{R_{in}+R_{out}}\\

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db

Analyzing SBAL:

Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099

the total gain will be the total gain of each stage multiplied by the load factor.

SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db

So the best amplifier arrangement is SABL.

4 0
3 years ago
A typical tire for a compact car is 22 inches in diameter. If the car is traveling at a speed of 60 mi/hr, find the number of re
Softa [21]

Answer:

rpm= 916.7436 rev/min

Explanation:

First determine the perimeter of the wheel, to know the horizontal distance it travels in a revolution:

perimeter= π×diameter= π × 22 inches × 0.0254(m/inche)= 1.7555m

Time we divide the speed of the car, which is the distance traveled horizontally over time unit, by the perimeter of the wheel that is the horizontal distance traveled in a revolution, this dates us the revolutions over the time unit:

revolutions per time= velocity/perimeter

velocity= (60 mi/hr) × (1609.34m/mi) = 96560m/h

revolutions per time= (96560.6m/h) / (1.7555m)= 55004.614 rev/hr

rpm= (55004.614 rev/hr) × (hr/60min)= 916.7436 rev/min

3 0
4 years ago
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